The geometrical quantity in damped wave equations on a square

October 2006, Vol. 12, 636–661 www.edpsciences.org/cocv

DOI: 10.1051/cocv:2006015

THE GEOMETRICAL QUANTITY IN DAMPED WAVE EQUATIONS ON A SQUARE

Pascal H´ ebrard

and Emmanuel Humbert

Abstract. The energy in a square membrane Ω subject to constant viscous damping on a subset ω ⊂ Ω decays exponentially in time as soon as ω satisfies a geometrical condition known as the

“Bardos-Lebeau-Rauch” condition. The rate τ(ω) of this decay satisfies τ(ω) = 2 min(−µ(ω), g(ω)) (see Lebeau [Math. Phys. Stud. 19(1996) 73–109]). Hereµ(ω) denotes the spectral abscissa of the damped wave equation operator andg(ω) is a number called the geometrical quantity ofωand defined as follows. A ray in Ω is the trajectory generated by the free motion of a mass-point in Ω subject to elastic reflections on the boundary. These reflections obey the law of geometrical optics. The geometrical quantity g(ω) is then defined as the upper limit (large time asymptotics) of the average trajectory length. We give here an algorithm to compute explicitlyg(ω) whenω is a finite union of squares.

Mathematics Subject Classification. 35L05, 93D15.

Received November 14, 2003. Revised July 19, 2004 and June 13, 2005.

Let Ω = [0,1]² be the unit square of R² and let ω ⊂Ω be a subdomain of Ω. We are interested here in the problem of uniform stabilization of solutions of the following equation

⎧⎪

⎪⎪

⎪⎨

⎪⎪

⎪⎪

⎩

utt(x, t)−∆u(x, t) + 2χω(x)ut(x, t) = 0, x∈Ω, t >^◦ 0, u(x, t) = 0, x∈∂Ω, t >0

u(.,0) =u₀∈H₀¹(Ω), ut(.,0) =u₁∈L²(Ω),

(0.1)

where χω is the characteristical function of the subset ω. This equation has its origin in a physical problem.

Consider a square membrane Ω. We study here the behaviour of a wave in Ω. Letu(x, t) be the vertical position of x ∈Ω at time t > 0. We assume that we apply on ω a force proportional to the speed of the membrane atx. Then,usatisfies equation (0.1). To get more information on this subject, one can refer to [2, 7, 10]. With regard to the one-dimensional case, the reader can consult [3, 6].

We do not deal here with the existence of solutions. We assume that there exists a solutionu. Let us define the energy ofuat time tby

E(t) =

Ω|∇u|²+u²_tdx.

Keywords and phrases. Damped wave equation, mathematical billards.

1Institut ´Elie Cartan, Universit´e de Nancy 1, BP 239 54506 Vandœuvre-l`es-Nancy Cedex, France;pascal [email protected];

[email protected]

c EDP Sciences, SMAI 2006

Article published by EDP Sciences and available at http://www.edpsciences.org/cocvor http://dx.doi.org/10.1051/cocv:2006015

It is well known that, for anyt≥0,

E(t)≤CE(0)e^{−2τ t}, (0.2)

whereC, τ ≥0. As proven by E. Zuazua [13], this result remains true in the semilinear case when the dissipation is in a neighborhood of a subset of the boundary satisfying the multiplier condition. Let us now define

Definition 0.1. The exponential rate of decayτ(ω) is defined by

τ(ω) = sup{τ≥0 s.t. ∃C >0 for which (0.2) holds}.

Many articles have been devoted to finding bounds forτ(ω). The reader can consult [1, 9, 11]. In the case of a non-constant damping, the reader may see [4]. Lebeau proved in [9] that

τ(ω) = 2 min(−µ(ω), g(ω)).

Hereµ(ω) denotes the spectral abscissa of the damped wave equation operator andg(ω) is a number called the geometrical quantity ofω and defined as follows. A ray in Ω is the trajectory generated by the free motion of a mass-point in Ω subject to elastic reflections in the boundary. These reflections obey the law of geometrical optics: the angle of incidence equals the angle of reflection. If a ray meets a corner of Ω, the reflection will be the limit of the reflection of the rays which go to this corner. It is easy to verify that the ray runs along the same trajectory before and after the reflection but in an opposite direction.

Letγ:R⁺→Ω be the parametrization of the ray by arclength. Let C=

rays in Ω .

Furthermore, let ρ₀ = (X₀, α), where X₀ ∈Ω andα∈[0,2π[. Consider the rayγρ₀ which starts atX₀ in the direction of the vector (cos(α),sin(α)). Each ray ofCcan be defined in this way. More precisely, if Γ = Ω×[0,2π[

then

C={γρ₀ s.t. ρ₀∈Γ}.

In the whole paperγ∈ Cwill be noted γ= [X₀, α]. Ifγ∈ C and ift >0 is a positive real number, we write:

γt=γ/[0,t]. Let

C =

t>0

γt γ∈ C

be the set of paths of finite length not necessarily closed. Forγt∈ C of lengtht >0, we define:

m(γt) =1 t

t

0 Xω(γt(s))ds.

The definition ofmcan be extended to C. Indeed, ifγ∈ C, set m(γ) = lim sup

t→+∞ m(γt).

It is proven in Section 1.2 that

m(γ) = lim

t→+∞m(γt). (0.3)

For a rayγ belonging toC,m(γ) represents the average time thatγ spends in Ω.

Definition 0.2. The geometrical quantityg(ω) is defined by g(ω) = lim sup

t→+∞ inf

γ∈Cm(γt).

We are interested here in a precise study of the geometrical quantityg. The first part of this paper is devoted to studying the rays. We recall some well known properties of rays. In the second part, another expression forg is given. Namely, we prove that ifω⊂Ω whose boundary is a finite union ofC¹-curves, then

g(ω) = inf

γ∈Clim sup

t→+∞ m(γt) = inf

γ∈Cm(γ).

It is easier to work with this second definition of g(ω). An important application of this theorem is given in the third part: we obtain an algorithm which gives an exact computation ofg(ω) when ω is a finite union of squares. This work has many interests. At first, Theorem 2.1 gives another definition for g(ω), much more easy to manipulate than the original one. Secondly, maximizing the exponential rate of decay is interesting from the point of view of physics. For these questions, knowing exactlyg(ω) is important. We give some exact computations ofg(ω) with the help of our algorithm at the end of the paper. An interesting question is then:

how can we choose the subsetω such that g(ω) is maximum? At the moment, this problem is still open. The following inequality is always trueg(ω)≤ |ω|(see Sect. 2) where|ω|is the area ofω. Let us set forα∈[0,1]

S(α) = sup

|ω|=αg(ω).

Some questions then arise naturally

– What is the value ofS(α)? In particular, is it true thatS(α) =α?

– Can we find a subsetω for whichg(ω) =|ω|?

– Can we find an optimal ω (i.e. anω that maximizesg) among the domains that satisfy the multiplier condition?

If |ω|= 0 or |ω|= 1, the answers are obvious. However, if |ω| ∈]0,1[, these questions seem to be much more difficult and are still open.

1. Basic properties of rays 1.1. Different representations of rays

In this section, we regard billards in Ω from different point of views (see for example [12]). Instead of reflecting the trajectory with respect to a side of Ω, one can reflect Ω with respect to this side. We then obtain a square grid and the initial trajectory is straightened to a line. This gives a correspondance between billard trajectories in Ω and straight lines in the plane equipped with a square grid. Two lines in the plane correspond to the same billard trajectory if they differ by a translation through a vector of the lattice 2Z+ 2Z. The factor 2 in 2Z+ 2Z is important. Indeed, two adjacent squares have an opposite orientation in the sense that they are symmetric with respect to their common side.

Now, consider T = [0,2]² and identify its opposite sides. A trajectory then becomes a geodesic line in the flat torus.

Finally, trajectories in Ω can be seen in three different ways

– the definition which consists in reflecting trajectories in sides of Ω;

– the point of view of straight lines inR²;

– the point of view of geodesic lines in a flat torusT.

Note that the set ω must be reflected in the same way as we did for Ω. This means that the setsω contained in two adjacent squares must be symmetric with respect to their common side (see Figs. 1 and 2).

X0

ω 1’=1

2’

3’

4’

0 1 2

1

Figure 1. Point of view of infinite plane.

X0

ω_T

0 1 2

1 2

Figure 2. Point of view of flat torus.

Lett >0 andγ∈ C. The value

1 t

t

0 Xω(γt(s))ds does not depend on the point of view we adopt.

1.2. Open and closed rays

A ray is said to be closed if it is periodic in the flat torus. It is said to be open if it is not closed. We have the following characterization. The ray [X₀= (x₀, y₀), α] is open if and only if cosαand sinαare independent over Z, i.e. if and only if tanα ∈ R−Q. Indeed, for example using the point of view of a flat torus, the trajectory is periodic if and only if there exist 0≤t₁< t₂ andn, m∈Nsuch that

x₀+t₁cosα=x₀+t₂cosα+ 2n y₀+t₁sinα=y₀+t₂sinα+ 2m.

Hence 2m(t₁−t₂) cosα−2n(t₁−t₂) sinα= 0. Let us now consider open rays. They have the following properties (see for example [5], p. 172).

Proposition 1.1. Letγ= [X₀, α] be an open ray. Then,{γt, t≥0} is dense in the torusT and then inΩ. In addition, ifω is quarrable (i.e. χω Riemann-integrable on Ω) then m(γt)−−−−→_t

→+∞ |ω|. More precisely, let αbe such that tanα∈R−Qthen

∀ >0,∃t₀|t≥t₀,∀X₀∈Ω : ifγρ₀ = [X₀, α]then |m(γt)− |ω||< . In other words,t₀ is independent ofX₀.

Here,|ω|stands for the area ofω. Consider now closed rays. They satisfy the following properties:

Proposition 1.2. Let γ= [X₀, α] be a closed ray. Then, there existp, q∈Z relatively prime such that:

cosα= p

p²+q² and sinα= q p²+q²· The periodL of the trajectory isL= 2

p²+q². We have m(γt)−−−−→_t

→+∞

1 L

L

0 χω(γ(s))ds.

Remark 1.1. The assertion (0.3) follows immediately from Propositions 1.1 and 1.2. Moreover, letγ∈ C. Ifγ is open, we havem(γ) =|ω|and ifγ is periodic of periodL, we havem(γ) =_L¹ L

0 χω(γ(s))ds.

As one can check, an immediate consequence of Proposition 1.2 is the following result Proposition 1.3. Let ω⊂Ωbe Riemann-integrable, then

g(ω)≤ |ω|.

Closed rays can easily be described. Let us adopt the point of view of a flat torus. Let us consider a ray γ = [X₀, α] with X₀ ∈T and cosα=p/

p²+q²,sinα=q/

p²+q², p andq relatively prime. We assume that p≥0 andq≥0 (i.e. α∈[0, π/2]). From Remark 3.1 below, the study can be restricted to such rays.

Theorem 1.1. The rayγ is an union of p+qparallel segments directed byu= cosαi+ sinαj. Among these segments, pof them start atAk,1≤k≤p, and theqothers start at Bk,1≤k≤qwith

Ak

0,

y₀+q/p(2k−x₀)−2E 1

2(y₀+q/p(2k−x₀))

and

Bk

x₀+p/q(2k−y₀)−2E 1

2(x₀+p/q(2k−y₀))

,0

.

In addition, the distance between two neighbouring segments is constant and equal to δ = 4/L, where L = 2

p²+q² is the period.

Proof. In the infinite plane R², the ray is defined by x(t) = x₀+ 2tp/L and y(t) = y₀+ 2tq/L, where L = 2

p²+q² is the period. Let us restrict the study to the segment [X₀X₁] of length L. In ]X₀X₁[, the ray intersects pvertical straight lines defined by the equationsx= 2k, k= 1, . . . , pandq horizontal straight lines defined by the equationy= 2k, k= 1, . . . , q. This givesp+qsegments starting at points with coordinates

0, y₀+q/p(2k−x₀)

X0

X1

∆ x = 2p

∆ y = 2q

A1

A2

A3

B1

B2

Ω

0 2 4 6 8

2 4 6

Figure 3. Closed ray forp= 3 andq= 2 in the infinite plane.

X₀ = X₁ A₁

A₂ A₃

B₁ B₂

δ δ δ δ

Ω

0 1 2

1 2

Figure 4. Closed rayp= 3 andq= 2 in the flat torus.

and

x₀+p/q(2k−y₀),0 .

Coming back to the point of view of the torus, this givesp+qsegments starting atAk andBk.

Sincepandqare relatively prime, it is well known that{kq[p],1≤k≤p}={0,1, . . . , p−1}. Hence, ifAk₁

andAk₂ are two neighbouring points,Ak₁Ak₂ = 2/p. As a consequence, the distance between the straight lines (Ak₁, u) and (Ak₂, u) is

δ= −−−−−→

Ak₁Ak₂.u^⊥ = 2

pj.−qi+pj p²+q² = 4

L·

In the same way, ifBk₁ andBk₂ are two neighbouring pointsBk₁Bk₂ = 2/q, the distance between (Bk₁, u) and (Bk₂, u) is

δ= −−−−−→

Bk₁Bk₂.u^⊥ = 2

qi. qi−pj p²+q² = 4

L·

Letk₁, k₂ be such that

dist(Ak₁,0) = min

k dist(Ak,0) anddist(Bk₂,0) = min

k dist(Bk,0).

It remains to prove that the distance between the two segments starting atAk₁ andBk₂ isδ= 4/L. Since the distance between two pointsAk is 2/p, there exists n∈Nsuch that the ordinatey^∗ofAk₁ satisfies

2n≤y₀+q/p(2k₁−x₀)−2E 1

2(y₀+q/p(2k₁−x₀))

<2n+ 2/p

=⇒ y^∗=y₀+q/p(2k₁−x₀)−2E 1

2(y₀+q/p(2k₁−x₀))

−2n <2/p.

In the same way, there existsm∈Nsuch that the abscissax^∗ ofBk₂ satisfies 2m≤x₀+p/q(2k₂−y₀)−2E

1

2(x₀+p/q(2k₂−y₀))

<2m+ 2/q

=⇒ x^∗ =x₀+p/q(2k₂−y₀)−2E 1

2(x₀+p/q(2k₂−y₀))

−2m <2/q.

The distanceδbetween the two segments verifies:

δ= −−−−−→

Ak₁Bk₂.u^⊥ = |qx^∗+py^∗| p²+q² ·

An easy computation shows that|qx^∗+py^∗|is an even integer number. Consequently,δcan be written as 4k/L, wherek is an integer. Since 0≤x^∗ <2/qand 0≤y^∗<2/p, we getδ <8/L, and hence,δ= 4/L.

Remark 1.2. If a quarrable domain ω is such that g(ω) = |ω| then for all α ∈ [0, π/2] and for almost all X₀∈T,m(γ) =|ω|, whereγ= [X₀, α].

Indeed, consider an open ray γ. Then m(γ) = |ω|. Let X₀ ∈ Ω. Assume that a ray γ starting at X₀ is closed of lengthL. LetM andN be two points ofγ such that−−→

M N is orthogonal touandM N =δ. Let also P(s) =sM+ (1−s)N andγ^s= [P(s), α], then

g(ω)≤ inf

s∈[0,1]m(γ_L^s)≤ ₁

0 m(γ_L^s)ds≤ |ω|.

If g(ω) = |ω|, then m(γ_L^s) = |ω| almost everywhere. If ∂ω is a finite union of curves of class C¹, then the number of discontinuity points of m(γ_L^s) is finite. It is null if ∂ω does not possess any segments. The remark above immediately follows.

2. An equivalent definition for g 2.1. The geometrical quantity g

We define, for any set ω:

Definition 2.1. The geometrical quantityg(ω) is defined by g(ω) = inf

γ∈Cm(γ).

As easily seen,g is easier to study thang. As shown in Section 3,g(ω) can be computed explicitly for a certain class of domainsω. Together with Theorem 2.1, this gives an algorithm for computingg(ω).

Theorem 2.1. Let ω be a closed set whose boundary is a finite union of curves of class C¹. Then g(ω) =g(ω).

As a first remark, the same argument as in the proof of Proposition 1.3 shows that

0≤g(ω)≤ |ω|. (2.1)

The proof of Theorem 2.1 is given in the appendix. It is easy to see thatg(ω)≤g(ω). Inequalityg(ω)≤g(ω) is much more difficult to obtain. To prove Theorem 2.1, we consider a sequence of rays γⁿ = [xn, θn] and a sequence of real numbers tn → +∞ for which m(γ_tⁿ_n) → g(ω). After choosing a subsequence, there exists (x, θ)∈Ω×[0,2π[ such that limnxn=xand limnθn =θ. We will now show that limnm(γ_tⁿ_n)−m([x, θ]t_n) = 0.

The conclusion then follows. The difficulty in this proof is that the function (x, θ)→m([x, θ]t) is not continuous.

A direct consequence of Theorem 2.1 is that it suffices to consider closed rays in the explicit computation ofg(ω).

Namely, letCc be the set of closed rays, then we have:

g(ω) = inf

γ∈Ccm(γ). (2.2)

3. Explicit computation of g

We prove in this section that for a particular class of domainsω, the properties of the geometrical quantity g we obtain above allow to compute explicitly g(ω). Let N ∈N^∗, and let ω ⊂Ω be a finite union of squares (Ci,j)_1≤i,j≤N, with

Ci,j = i−1

N , i N

× j−1

N , j N

·

Obviously,ω is Riemann-measurable, closed and its boundary is a finite union ofC¹ curves.

3.1. Influence of α

The first result we obtain is the following

Theorem 3.1. Let γ = [X₀, α] be a closed ray such that tanα = q/p, with p and q relatively prime and p+q >2N, then

|m(γ)− |ω| | ≤N²

pq min{|ω|,1− |ω|} ≤ N² 2pq·

Proof. We adopt the point of view of flat torusT = [0,2]². At first, letK be one of the Ci,j i.e.

K=Ci₀,j₀ =

i₀−1 N ,i₀

N

×

j₀−1 N ,j₀

N

=ABCD.

The period of γ is L= 2

p²+q². Letδ = 4/L. Then cosα=pδ/2 and sinα=qδ/2. By Theorem 1.1, the rayγis constituted byp+qsegments directed byu. The distance between two neighbouring segments isδ.

Let us define the orthonormal frame Ru = (X,−→ Iu,−→

Ju), where X is such that −−→

XD.u = 0, −−→

XD.−−→

XA = 0,

−

→Iu=−−→XD/||−−→XD||, and−→

Ju=−−→XA/||−−→XA||=u(see Fig. 5).

In this frame, the coordinates ofA, B, C and Dsatisfy A

0,sinα

N

; B sinα

N ,sinα+ cosα N

; C

sinα+ cosα N ,cosα

N

; D

cosα N ,0

.

A B

D C X

0 1 2

1 2

Figure 5. Rayγ= [0, α] with tanα= 4/15.

A B

C

D

X x

0 x

1 x

2

Figure 6. Vertical segments in the squareABCD.

The ray γ is constituted ofp+qvertical segments. R of them meetK (see Fig. 6). We note x₀, x₁, . . . xR−1

their abscissa

x₀=r∈[0, δ]

xi=x₀+iδ, 0≤i≤R−1.

Since ^sinα+cos_N ^α= ^p₂⁺_N^qδ > δ, there exists at least one segment inK andR≥1.

Let

f₁(x) = cotα x+sinα

N = cotα

x−sinα N

+sinα+ cosα

N line (AB)

f₂(x) =−tanα

x−sinα N

+sinα+ cosα N

=−tanα

x−sinα+ cosα N

+cosα

N line (BC) g₁(x) =−tanα x+sinα

N =−tanα

x−cosα N

line (AD) g₂(x) = cotα

x−cosα N

= cotα

x−sinα+ cosα N

+cosα

N line (CD) and

f(x) =

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎨

⎪⎪

⎪⎪

⎪⎪

⎪⎩ sinα

N ifx≤0 f₁(x) if 0≤x≤ ^sin_N^α

f₂(x) if ^sin_N^α ≤x≤^sinα+cos_N ^α cosα

N ifx≥^sinα+cos_N ^α

andg(x) =

⎧⎪

⎪⎪

⎪⎪

⎪⎨

⎪⎪

⎪⎪

⎪⎪

⎩ cosα

N ifx≤0 g₁(x) if 0≤x≤ ^cos_N^α

g₂(x) if ^cos_N^α ≤x≤ ^sinα+cos_N ^α cosα

N ifx≥ ^sinα+cos_N ^α·

Then

Rf−g=

^sinα+cosα

N

0 (f(x)−g(x))dx=|K|= 1 N² and

1 L

L

0 χK(γ(s))ds=δ 4

R−1 i=0

f(xi)−g(xi).

Letkbe the number of vertical segments which meet [AB], i.e. thek segments such that r+ (k−1)δ≤sinα

N ≤r+kδ.

If 1≤k≤R−1, if at least one segment meets [AB] and if at least one segment meets [BC] then, as one can see

1 L

R−1 i=0

f(xi) = δ 4

k−1

i=0

f₁(xi) +δ 4

R−1 i=k

f₂(xi)

= 1 4

k−1

i=0

x_i+δ/2

x_i−δ/2 f₁(x)dx+1 4

R−1 i=k

x_i+δ/2

x_i−δ/2 f₂(x)dx

= 1 4

r+(k−1/2)δ

r−δ/2 f₁(x)dx+1 4

r+(R−1/2)δ

r+(k−1/2)δ f₂(x)dx. (3.1) Ifk= 0 ork=R, equation (3.1) remains true. Hence, in all cases

1 L

R−1 i=0

f(xi) =1 4

₍p+q)δ/2N 0

f(x)dx−1 4

r−δ/2 0

f₁(x)dx

+1 4

r+(k−1/2)δ

qδ/2N (f₁(x)−f₂(x))dx+1 4

r+(R−1/2)δ

(p+q)δ/2N f₂(x)dx.

In the same way, letl be the number of segments which meet [AD]

r+ (l−1)δ≤ cosα

N ≤r+lδ.

We obtain that, in all cases 1

L

R−1 i=0

g(xi) =1 4

r+(l−1/2)δ

r−δ/2 g₁(x)dx+1 4

r+(R−1/2)δ

r+(l−1/2)δ g₂(x)dx

=1 4

₍p+q)δ/2N

0 g(x)dx−1 4

r−δ/2

0 g₁(x)dx +1

4

r+(l−1/2)δ

pδ/2N (g₁(x)−g₂(x))dx+1 4

r+(R−1/2)δ

(p+q)δ/2N g₂(x)dx.

Finally 1 L

L

0 χK(γ(s))ds− 1

4N² =−1 4

r−δ/2

0 f₁−g₁+1 4

r+(k−1/2)δ

qδ/2N f₁−f₂ +1

4

r+(l−1/2)δ pδ/2N

g₁−g₂+1 4

r+(R−1/2)δ (p+q)δ/2N

f₂−g₂

=tanα+ cotα 8

−(r−δ/2)²+ (r+ (k−1/2)δ−qδ/2N)²

+(r+ (l−1/2)δ−pδ/2N)²−(r+ (R−1/2)δ−(p+q)δ/2N)² .

However, in this last equality, each of the four square terms is less than δ/2 and tanα+ cotα= _sinα¹_cosα =

pqδ4² =₄^L_pq² hence

1 L

L

0 χK(γ(s))ds− 1 4N²

≤ 1

4pq· (3.2)

We now consider:

ωT =

(i,j)∈E

Ci,j,

whereE is a subset of {1,2, . . . ,2N}²and card(E) =N²|ωT|= 4N²|ω|; then m(γ) = 1

L L

0 χω_T(γ(s))ds=

(i,j)∈E

1 L

L

0 χC_i,j(γ(s))ds.

Hence

m(γ)− |ω|=

(i,j)∈E

1 L

L

0 χC_i,j(γ(s))ds− 1 4N²

· This leads to

|m(γ)− |ω| | ≤card(E) 4pq =N²

pq|ω|.

If the area ofωis greater than 1/2, letω be the adherence of Ω−ωwhich is also a finite union of squaresCi,j.

Hence

1 L

L

0 χω(γ(s))ds−(1− |ω|) ≤ N²

pq(1− |ω|)

0 1 C1

C 2 2

C 3 3

C 4 4

C 5 5

C 6 6 L 1

1 L₂2 L 3

3 L 4

4 L 5

5 L 6

6

columns

lines

x y

Figure 7. A domainω forN = 6.

or

1 L

L

0 χω(γ(s))ds= 1− 1 L

L

0 χω(γ(s))ds.

Finally, we get|m(γ)− |ω| | ≤ ^N_pq²(1− |ω|).

3.2. Representation of ω

For convenience, consider the frame (O, I , J) withI=i/(2N) andJ=j/(2N). As a consequence, Ω is now the square [0, N]², andT is the torus [0,2N]². The length of closed rays defined by (p, q) (p, qrelatively prime) such that cosα=p/

p²+q² and sinα=q/

p²+q² must be modified. The period is now L= 2N

p²+q² andCi,j is such that:

Ci,j= [i−1, i]×[j−1, j].

Let us define the matrixM N×N such thatMi,j= 1 ifCj,i⊂ω, andMi,j= 0 ifCj,i⊂ω. With this notation, the matrix M is a representation ofω. As an example, ifω is as in Figure 7 then

M =

⎛

⎜⎜

⎜⎜

⎜⎜

⎝

1 1 0 0 1 1

1 1 1 1 0 1

0 1 0 1 0 1

0 0 0 0 1 1

1 1 0 0 0 0

0 1 1 0 0 0

⎞

⎟⎟

⎟⎟

⎟⎟

⎠ .

Let us also define the 2N×2N matrixAsuch thatAi,j = 1 ifCj,i⊂ωT, Ai,j= 0 ifCj,i⊂ωT,i.e.

Mi,j =Ai,j =A₂N+1−i,j=Ai,2N+1−j=A₂N+1−i,2N+1−j. (3.3)

X0

X₁ X₂

X3

u u₁ u₂

u₃

u u

u

0 1 2

1 2

Figure 8. Starting points withα∈[0, π/2].

3.3. Influent rays

Letγ be a horizontal ray starting atX₀= (x₀, y₀). Then, m(γ) = 1

N N j=1

Mi,j.

Assume thaty₀=kis an integer. Since eachCi,j is closed, m(γ) = 1

N N j=1

(Mk−1,j+Mk,j).

Obviously, this ray does not realize the infimum in the definition ofg. In the same way, for a vertical ray, only the numbers

1 N

N i=1

Mi,j, 1≤i≤N

must be considered. Now, let us deal with oblique rays and fix α. Let us denote tanαbyq/p, wherepand q are relatively prime integers.

Remark 3.1. One has to consider only the rays defined by γ = [X₀, α] with X₀ ∈ T and α∈ [0, π/2], i.e.

p≥0 andq≥0.

Indeed, keeping the notations of Figure 8, the rayγ¹ = [X₀,−u→₁] is equivalent to ˜γ¹ = [X₁, u]. In the same way, the raysγ²= [X₀,−→u₂] andγ₃= [X₀,−→u₃] are equivalent to ˜γ²= [X₂, u] and ˜γ³= [X₃, u].

Let us now study the influence of the starting point of rays. We have the following result

Proposition 3.1. Among oblique closed rays of angleα, the ray which spends the least time inω in average is a ray which meets a point with integer coordinates.

Proof. Let B = (x₀, y₀) be a point of T such that the ray γ = [B, α] does not meet a point with integer coordinates. Let us adopt the point of view of infinite plane and let B₀, B₁, B₂, . . . be the intersection points of γ with the lattice Z+Z. A direct consequence of Theorem 1.1 is that B₀B₂N(p+q) = L = 2N

p²+q².

u= (pi+qj)/

p²+q² is the direction of the ray. Letn= (−qi+pj)/

p²+q² be such that (u, n) is a direct

B₀

B₁ B₂

B₃

B₄ B₅

B₆

B₇ B₈

B₉

B₁₀ B₁₁

B₁₂

A₀

A₁ = A₂

A₃

A₄ = A₅

A₆

A₇ = A₈

A₉

A₁₀ = A₁₁

A₁₂

C₀

C₁

C₂ = C₃

C₄

C₅ = C₆

C₇

C₈ = C₉

C₁₀

C₁₁ = C₁₂

0 2 4 6 8

2 4

Figure 9. Case of a ray which does not meet any point with integer coordinates.

orthonormal frame. For a point P = (k, l) with integer coordinates, let us compute the algebraic distance to the rayγ

d(P, γ) =−−→

BP .n=pl−qk−(py₀−qx₀) p²+q² · There exists a pointAwith integer coordinates and which verifies:

d(A, γ) = min{d(P, γ)|P has integer coordinates andd(P, γ)>0}.

Let Ai be the intersection point of the line meeting A and directed by u(the points A solutions of the mini- mization problem above are clearly on a same line directed by u) and the line belonging to the latticeZ+Z which contains Bi (see Fig. 9 for the case N = 2). In the same way, there existsC with integer coordinates which satisfies

d(C, γ) = min{d(P, γ)|P has integer coordinates andd(P, γ)<0}.

Let us note Ci the intersection of the line meeting C and directed byuand the line belonging to the lattice Z+Zwhich contains Bi (see Fig. 9).

There are three oblique rays of angleα γA= [A₀, α], γB = [B, α] = [B₀, α] andγC= [C₀, α]. According to the definitions ofAandC, the segments [AiAi+1], [BiBi+1] and [CiCi+1] belong to the same squareCk,l. We defineεk = 1 if [BkBk+1]⊂ωT,εk= 0 if [BkBk+1]⊂ωT. Then, if

m(γB) = 1 L

[B₀B_2N(p+q)]χω_T = 1 L

2N(p+q) k=1

εkBk−1Bk,

we have:

m(γA) = 1 L

2N(p+q) k=1

εkAk−1Ak and m(γC) = 1 L

2N(p+q) k=1

εkCk−1Ck.

These three rays are parallel. Hence, there existst∈(0,1) such that for alli,Biis the barycenter of (Ai, t) and (Ci,1−t). If the quadrilateralAiAi+1CiCi+1is a trapezium (or in the degenerate case a rectangular triangle), Thales’ theorem implies thatBiBi+1 =t AiAi+1+ (1−t)CiCi+1. If the quadrilateral is a parallelogram, this inequality remains valid because the three lengths are equal. As a consequence, we have

m(γB) =tm(γA) + (1−t)m(γC).

The minimum of the three quantities m(γA), m(γB) andm(γC) is attained for one of the two raysγA ouγC. This shows that γcannot be the minimum in the set of oblique closed rays.

It remains to find the anglesαwe need to consider among the rays which meet a point with integer coordinates.

In the flat torusT = [0,2N]², there are 4N² possible starting points. However, the rayγ = [(x₀, y₀), α] with (x₀, y₀)∈N² meets 2N points with integer coordinates. Indeed, γmeets the point (x₁, y₁) if and only if there existsλ∈[0,2N) such that

x₁=x₀+λp [2]

y₁=y₀+λq [2].

Therefore, the points belonging to γ and with integer coordinates are exactly the 2N points obtained for λ= 0,1, . . . ,2N−1. There exist at least 2N rays which meet a point with integer coordinates.

Proposition 3.2. Let dbe the G.C.D. ofpand2N, let d be such thatd d = 2N. Then, the2N rays of angle αstarting at (i, j)with 0≤i≤d−1 and0≤j≤d−1 meet one and only one time all the points of T with integer coordinates.

Proof. Letp be such thatd p=p. We note thatp andd are relatively prime. Let alsoγ₁andγ₂ be the rays starting respectively at A and B with integer coordinates (k₁, k₂) and (k₃, k₄) (with 0 ≤k₁, k₃ ≤d−1 and 0≤k₂, k₄≤d−1 ). There exists a point C∈γ₁∩γ₂ with integer coordinates if and only if there exist four integer numbersλ, µ, nandmsuch that

x=k₁+λp=k₃+µp+ 2N n y=k₂+λq=k₄+µq+ 2N m.

Therefore,k₁−k₃= (µ−λ)p+ 2N nandk₁−k₃is divisible byd. Since−(d−1)≤k₁−k₃ ≤d−1, we have k₁=k₃ and (µ−λ)p=−2N n. As a consequence, (µ−λ)p =−dnand since p andd are relatively prime, (µ−λ) =kd, where k∈ Z. In the same way, k₂−k₄ = (µ−λ)q+ 2N m=kdq+ 2N m. Hence k₂−k₄ is divisible byd. Since−(d−1)≤k₂−k₄≤d−1, we havek₂=k₄ andγ₁=γ₂.

3.4. The algorithm

In this section, we give a method to compute explicitlyg when ω is a finite union of squaresCi,j. At first, according to Propositions 1.1 and 3.1, the study can be restricted to closed rays starting at a point with integer coordinates.

The user of the program must input the matrixM which representsωand the value of a parameterP Qmax which avoids infinite loops (this problem never appeared until now).

The algorithm is the following

1. Compute the minimum gof the 2N numbers 1

N N i=1

Mi,j and 1 N

N j=1

Mi,j,

which corresponds to the minimum ofm(γ) whenγ is a vertical or horizontal ray.

2. From the matrixM, build the matrixA defined in (3.3).

3. For all (p, q) such thatp≥1,q≥1,p+q≤2N andpandqrelatively prime, compute the number (the way to computemp,q is explained below)

mp,q = min

X∈{1,2,...,2N}²{m(γ)|γ= [X,arctanq/p]}. (3.4) Theng←min{g, mp,q}.

4. If at this stepg=|ω|, the user inputs the parameterP Qmaxwhich corresponds to the greatest product pq which will be considered; in the other casesP Qmax=E(_|ω^N_|−²_gmin{|ω|,1− |ω|}) + 1.

5. Find all the (p, q) relatively prime such thatp≥1,q≥1,p+q >2N,pq≤P Qmax. Sort these couples from the lowest productpq to the greatest one. This gives a listLcontainingncouples.

6. i←1 and as long as|ω| −g≥ ^N_pq² min{|ω|,1− |ω|}andi≤n, g←min{g, mp,q} and i←i+ 1,

where (p, q) is thei-thcouple ofLandmp,q is defined above (see (3.4)).

7. Finally

• if at the end of this loop,g=|ω|, then

|ω| − N²

P Qmaxmin{|ω|,1− |ω|} ≤g(ω)≤ |ω|.

This means that one of the following cases occurs;g(ω) =|ω|, or too few families of rays have been considered;

• ifg=|ω|, according to Theorem 3.1,g(ω) is equal tog.

We now present the method of computation defined in (3.4) for two relatively prime integers p≥ 1, q ≥1.

At first, let us consider the rayγ₀ = [0,arctanq/p] and let us study the way to compute m(γ₀). Its period is L= 2N

p²+q²and between the instantst= 0 andt=L, it meets 2N(p+q−1) + 1 points which have at least one integer coordinate. Let P₁, . . . , P₂N(p+q−1)+1 be these points (see Fig. 10). Remember that Pl(p+q−1)+1, 1≤l≤2N have integer coordinates.

Let (ik), (jk) and (εk) be the three other finite sequences (1≤k≤2N(p+q−1)) defined by PkPk+1⊂[ik, ik+ 1]×[jk, jk+ 1]

and

εk=

1 if [ik, ik+ 1]×[jk, jk+ 1]⊂ω 0 if not.

Then, m(γ₀) = _L^{1 2}_k^N₌₁^(p+q−1)εkPkPk+1, the computation of m(γ₀) is equivalent to the computation of the sequences (PkPk+1), (ik) and (jk).

• Computation of the lengthsPkPk+1

At first, note that it suffices to compute the lengthsPkPk+1for 1≤k≤p+q−1. Indeed,−−−−−−−−−→

P_k+(p+q−1)Pk=pi+qj,